I want to assume that $A$ is in Jordan canonical form with all of the zero eigenvalues grouped at the lower/right part of the main diagonal, and then let $U$ be the "backwards-identity" matrix ($ij$th entry is 0 unless $i+j=n+1$, in which case it is 1), so that conjugation by $U$ is equivalent to rotating the matrix 180${}^\circ$, and set $B=U^{-1}AU$. Morally speaking, this wants to be a rank-$n$ matrix. Obstacles: one might accidentally have two eigenvalues add to 0 on the main diagonal; the resulting matrix has nonzero entries just above and below the main diagonal - still invertible?
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Greg MartinAug 4 '12 at 2:11

Greg, what is JNF when K is not algebraically closed?
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Yemon ChoiAug 4 '12 at 3:55

Greg, surely a modification of this works, if instead of insisting on rotating the matrix 180 degrees, you allow yourself a more general permutation of the Jordan blocks. This should take care of the algebraically closed case, and I don't see how to do the general case.
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Aaron TikuisisAug 4 '12 at 18:10

1 Answer
1

Initially my assumption was that the question was primarily about finite fields $K$, but after reading the comments it appears to me that there is some confusion about the infinite case as well. So I'll prove the claim for $|K|=\infty$. I would guess the assertion is also true if $K$ is finite, but I don't know how to show it.

If one assumes $K$ to be algebraically closed, then it's very easy to see that one can find an appropriate $B$: Let $\lambda_1,\ldots,\lambda_n$ be the eigenvalues of $A$ (eigenvalues with multiplicity $m$ are supposed to appear $m$ times in that list), then for any permutation $\sigma\in S_n$ we can conjugate $A$ to a matrix of the form
$$
\textrm{diag}(\lambda_{\sigma(1)}, \ldots, \lambda_{\sigma(n)}) + N
$$
where $N$ is a strictly upper triangular matrix.
(This should be clear but I'll give a short explantion anyhow: View $K^n$ as a $K[x]$ module where $x$ acts as $A$. Then finding a basis that conjugates $A$ into the form given above corresponds to finding a filtration $0= V_0 < V_1 <\ldots< V_n= K^n$ of the $K[x]$-module $K^n$ such that the quotient $V_i/V_{i-1}$ is isomorphic to the simple $K[x]$-module $K[x]/(x-\lambda_{\sigma(i)})$. But by the structure theorem on f. g. modules over PID's, $K^n$ is a direct sum of modules each having just one isomorphism type of simple modules as composition factors. So choosing such a filtration is no problem: in each step just pick a maximal submodule in the desired isotypic direct summand and leave the other summands unchanged.)

We may assume that $A$ is of the above shape with $\sigma=\textrm{id}$. So just choose a permutation $\sigma$ such that $\lambda_i+\lambda_{\sigma(i)} \neq 0$ for all $i$ (this can be done: just choose $\sigma$ to be the transposition that swaps each $\lambda_i$ which is zero with some $\lambda_i$ which is non-zero and fixes the rest; the rank condition implies that there are enough non-zero eigenvalues to pair with the zero eigenvalues; also $char(K)\neq 2$ is essential: it guarantees $\lambda_i\+\lambda_i\neq 0$ whenever $\lambda_i\neq 0$). Then $A+B$ can becomes
$$
\textrm{diag}(\lambda_1+\lambda_{\sigma(1)}, \ldots, \lambda_n+\lambda_{\sigma(n)}) + \textrm{something strictly upper triangular}
$$
and you are done.

As for the case of a non-algebraically closed (but still infinite) $K$: Note that the algebraically closed case implies that the rational function
$$
det(A+TAT^{-1})=\frac{det(AT+TA)}{det(T)} \in K(T_{ij})
$$
is non-zero. But $GL(n,K)$ is Zariski dense in $GL(n,\bar K)$ (see here), and so the above rational funktion must be non-vanishing on $GL(n,K)$, which shows there is a $T\in GL(n,K)$ such that $det(A+TAT^{-1})\neq 0$.

Thanks Florian. Necessarily $char(K)\not=2$. Else, consider the matrix $A=diag(I_p,0_{n-p})$ where $p>\dfrac{n}{2}$. Then there are no matrices $X$ s.t. $AX+XA$ is invertible. I have a question about the Zariski-density. In our instance, the group is reductive but is it connected ? It remains to show the result when $K$ is a finite field. Does there exist an obvious obstruction ?
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loup blancAug 7 '12 at 4:26

That $GL(n,\overline{K})$ is connected follows by identifying it with a closed subvariety of $\mathbb{A}^{n^2 + 1}$ rather than as an open subvariety of $\mathbb{A}^{n^2}$ since as a closed subvariety, it can be seen as the one defined by the vanishing of the ideal generated by $\textrm{det}T - 1$ where $T$ is the "extra" variable in the polynomial ring. That this is a prime ideal just means that it is an irreducible polynomial, which is an easy exercise (it has almost nothing to do with how the determinant looks).
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Tobias KildetoftAug 7 '12 at 6:33

1

Hi Florian, I reread your proof and unfortunately it is false. Indeed your assertion: "the rank condition implies that there are enough non-zero eigenvalues to pair with the zero eigenvalues" is not true. In particular, your proof does not work if $A$ is a nilpotent matrix. The Greg's proof does not work any more; choose, for instance, A=JordanBlockMatrix([[0, 4], [0, 2], [0, 3]]) (according to the maple notation). In fact, the asked result is true for the previous matrix. To the moderator (S. Carnahan): I'd want to edit a question (to Tobias) in this file. I don't know how to do. Thanks.
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loup blancAug 15 '12 at 21:22